Accelerated frame approximation in Schwarzschild metric far from the horizon It is clear to me that if I take the Schwarzschild metric $$ds^2 = \left(1-\frac{2M}{r}\right)dt^2 - \left(1-\frac{2M}{r}\right)^{-1} dr^2$$ and choose $\rho = 2\sqrt{\frac{r}{2M} -1}$ then I get the Rindler coordinates when $\rho << 1$.$$ ds^2 = \frac{\rho^2}{4} dt^2 - 4M^2 d\rho^2$$ Therefore I can say that when I am close to the horizon of a black hole at constant $\rho$ (or $r$ equivalently) then I am constantly accelerating outwards and do my space-time diagrams in Rindler coordinates. But this approximation is not valid when $\rho >> 1$, i.e. when the field is weak.
In the weak field approximation we have $$ds^2 = \left(1-\frac{2M}{r}\right)dt^2 - \left(1+\frac{2M}{r}\right) dr^2$$ my intuition tells me that to stay at $r$ constant I have to be constantly accelerating outwards, but I can't find any coordinate transform that brings me to the Rindler coordinates from the weak field approximation.
Please help :)
 A: Before looking for a solution, we should think carefully about what kind of result we should expect. A metric of the form
$$
   \rho^2 dt^2- d\rho^2
\tag{1}
$$
represents flat spacetime, and any worldline with constant $\rho$ is undergoing constant acceleration (constant weight). Because it is flat, a metric of this form can't be a good approximation to the Schwarzschild metric for all $\rho$. We can only expect it to be a good approximation for $\rho=\rho_0+\epsilon$ for some given $\rho_0$ and with sufficiently small $\epsilon$. Near the horizon, we have such an approximation with $\rho_0=0$; the hovering observer's acceleration diverges at the horizon, as it does in flat spacetime at the Rindler horizon. But away from the horizon, we will have $\rho_0\neq 0$, and then we can only expect a flat-spacetime approximation (1) to be valid modulo terms of order $\epsilon^2$. 
With that in mind, start with the Schwarzschild metric in coordinates $t$ and $\mathbf{x}\equiv (x,y,z)$:
$$
d\tau^2=A(r)dt^2-\frac{dr^2}{A(r)}-(d\mathbf{x}^2-dr^2)
\\
r\equiv\sqrt{x^2+y^2+z^2}
\hskip2cm
A(r)\equiv 1-\frac{2M}{r}.
\tag{3}
$$
Let $\mathbf{x}=(0,0,z_0)$ be the worldline of the hovering observer, and expand everything to first order in $x$, $y$, and $\delta z\equiv z-z_0$. This gives
$$
r^2\approx z_0^2+2z_0\delta z
\hskip1cm
\Rightarrow
\hskip1cm
r\approx z_0+\delta z=z
\tag{4}
$$
and
$$
A(r)\approx 1-\frac{2M}{z}\approx A(r_0)+\frac{2M}{z_0^2}\delta z
\equiv a+bz.
\tag{5}
$$
All approximations are understood to be valid modulo terms of order $(\delta z)^2$. Use these in (3) to get
$$
d\tau^2\approx (a+bz)dt^2-\frac{dz^2}{a+bz}-dx^2-dy^2.
\tag{6}
$$
Now define $\rho$ by
$$
\rho^2\equiv a+bz
\tag{7}
$$
to get the Rindler-like form
$$
d\tau^2\approx \rho^2 dt^2-\frac{d\rho^2}{(b/2)^2}-dx^2-dy^2,
\tag{8}
$$
with the understanding that this is only valid in a neighborhood of $\mathbf{x}= (0,0,z_0)$, which corresponds to $\rho_0=\sqrt{a+bz_0}$. The approximation is good modulo terms of order $\epsilon^2$, where $\epsilon\equiv\rho-\rho_0$. This is the desired result.
